Question

Using the following equation, find the center and radius of the circle. You must show all work and calculations to receive credit.

x2 −2x + y2 − 6y = 26

Answer 1

well you need to complete the square in both x and y

so you have

\[(x^2 -2x + ?) + (y^2 -6y + ??) = 26+ ? + ??\]

are you able to complete the square...?

Answer 2

No I really have no idea what to do

Answer 3

ok...

can you think of a square number where two of the factors add to -2....?

Answer 4

the square numbers

1, 4, 9, 16, 25, .....

so you need to add the correct square number to each part of the equation

Answer 5

-1 and 0

Answer 6

well close.... 1 because -1 x -1 = 1

so you now have

\[(x^2 - 2x + 1) + (y^2 - 6y + ??) = 26 + 1 + ??\]

you need to add the value to both sides of the equation to keep it in balance

so now look at y...

what square number has factors that would add to -6...?

Answer 7

3

Answer 8

well 3^2 = 9

so its

\[(x^2 - 2x + 1) + (y^2 - 6x + 9) = 26 + 1 + 9\]

now you need to complete factor both quadratics

\[(x - 1)^2 + (x - 3)^2 = 6^2\]

so the general form is

\[(x - h)^2 + (y - k)^2 = r^2\]

so the centre is (h, k) and radius is r

you need to match this to your factored equation to find the centre and radius

Answer 9

How do I do that sorry I'm really bad at math

Answer 10

well if you match

\[(x -h)^2 + (y - k)^2 = r^2 ...with...(x - 1)^2 + (y - 3)^2 = 6^2\]

h = 1, k = 3 so the centre is (1, 3) and the radius is 6

Answer 11

Is that all I need

Answer 12

that's the final answer for the centre and radius...

Answer 13

Thank you